Determine all integers m for which a square of length m can be dissected into five rectangles, the side lengths of which are the integers 1,2,3,...,10 in some order.
The smallest possible for the sum of the 5 areas is
10*1+9*2+8*3+7*4+6*5=110
and the largest sum is
10*9+8*7+6*5+4*3+2*1=190
So the square must have sides of length 11, 12, or 13.
The only way to dissect a square into 5 rectangles is to have one in the center surrounded by the other 4.
If the square has side 13, the consecutive 4 outer rectangles must have pairs of length and width that sum to 13. The only pairs that sum to 13 are 10+3, 9+4, 8+5, 7+6 which leaves the center as a 1*2 rectangle. The two pairs opposite rectangles must leave gaps of 1 and 2 so must sum to 12 and 11. There are not a pair of pairs that can do this,. (For example 8+3=11 but that only leaves the numbers in the pairs 9,4 and 7,6 and there's no sum to 12.)
If the square has side 12, the 4 outer rectangles have pairs with um 12. The only pairs with sum of 12 are 10+2, 9+3, 8+4, 7+5 which leaves the center as a 1*6 rectangles. The gaps this time would be 11 and 6. Again, no pair of pairs works.
There is at least one solution for a 11x11 rectangle but I don't have time top enter my analysis.
|
|
Posted by Jer
on 2020-10-02 13:06:39 |
Solution part 1
